# A triangle has corners at (2 , 2 ), ( 5, 6 ), and ( 1, 4 ). What are the endpoints and lengths of the triangle's perpendicular bisectors?

Jan 15, 2016

Each of the 3 segments have one endpoint at the midpoint of the given points and the other endpoint at $\left(2.5 , 4.75\right) , \left(3.5 , 4\right) \mathmr{and} \left(\frac{53}{22} , \frac{28}{11}\right)$, and lengths equal to $1.25 , .5 \cdot \sqrt{5} \mathmr{and} 5 \cdot \frac{\sqrt{5}}{11}$.

#### Explanation:

Repeating the points
$A \left(2 , 2\right) , B \left(5 , 6\right) , C \left(1 , 4\right)$

Midpoints
${M}_{A B} \left(3.5 , 4\right)$, ${M}_{B C} \left(3 , 5\right)$, ${M}_{C A} \left(1.5 , 3\right)$

Slopes of segments ($k = \frac{\Delta y}{\Delta x}$, $p = - \frac{1}{k}$)
$A B \to {k}_{1} = \frac{6 - 2}{5 - 2} = \frac{4}{3} \to {p}_{1} = - \frac{3}{4}$
$B C \to {k}_{2} = \frac{4 - 6}{1 - 5} = \frac{- 2}{-} 4 = \frac{1}{2} \to {p}_{2} = - 2$
$C A \to {k}_{3} = \frac{4 - 2}{1 - 2} = \frac{4}{- 1} = - 2 \to {p}_{3} = - \frac{1}{2}$

Now to simplify the work that remains to do, be noted that the line perpendicular to a side meets one of the other sides (or both at the same time when it intercepts a vertex of the triangle). When it meets just one of the two other sides, which side is this? Answer: the LONG side (although that is intuitive it can be proved). Then we need to know the lengths of the sides of the triangle.
$A B = \sqrt{{\left(5 - 2\right)}^{2} + {\left(6 - 2\right)}^{2}} = \sqrt{9 + 16} = 5$
$B C = \sqrt{{\left(1 - 5\right)}^{2} + {\left(4 - 5\right)}^{2}} = \sqrt{16 + 4} = 2 \cdot \sqrt{5} \cong 4.472$
$C A = \sqrt{{\left(1 - 2\right)}^{2} + {\left(4 - 2\right)}^{2}} = \sqrt{1 + 4} = \sqrt{5} \cong 2.236$
=> $A B > B C > C A$

So
line 1 perpendicular to AB meets side BC
line 2 perpendicular to BC meets side AB
line 3 perpendicular to AC meets side AB

We need the equations of the lines in which the sides AB and BC lay and the equations of the 3 perpendicular lines

Equation of the line that supports side:
$A B \to \left(y - 2\right) = \left(\frac{4}{3}\right) \left(x - 2\right)$ => $y = \frac{4 x - 8}{3} + 2$ => $y = \frac{4 x - 2}{3}$ [a]
$B C \to y - 4 = \left(\frac{1}{2}\right) \left(x - 1\right)$=>$y = \frac{x - 1}{2} + 4$=>$y = \frac{x + 7}{2}$[b]

Equation of the line (passing through midpoint) perpendicular to side:
$A B \to \left(y - 4\right) = \left(- \frac{3}{4}\right) \left(x - 3.5\right)$ => $y = \frac{- 3 x + 10.5}{4} + 4$ => $y = \frac{- 3 x + 26.5}{4}$ 
$B C \to \left(y - 5\right) = - 2 \left(x - 3\right)$ => $y = - 2 x + 6 + 5$ => $y = - 2 x + 11$ 
$A C \to \left(y - 3\right) = - \left(\frac{1}{2}\right) \left(x - 1.5\right)$ => $y = \frac{- x + 1.5}{2} + 3$ => $y = \frac{- x + 7.5}{2}$ 

Finding the interceptions on sides AB and BC

Combining equations [b] and 

{y=(x+7)/2
{y=(-3x+26.5)/4 => $\frac{x + 7}{2} = \frac{- 3 x + 26.5}{4}$ => $2 x + 14 = - 3 x + 26.5$ => $5 x = 12.5$ => $x = 2.5$
$\to y = \frac{2.5 + 7}{2}$ => $y = 4.75$

We've found $R \left(2.5 , 4.75\right)$
The distance between ${M}_{A B}$ and R is
$d 1 = \sqrt{{\left(2.5 - 3.5\right)}^{2} + {\left(4.75 - 4\right)}^{2}} = \sqrt{1 + .5625} = 1.25$

Combining equations [a] and 

{y=(4x-2)/3
{y=-2x+11 => $\frac{4 x - 2}{3} = - 2 x + 11$ => $4 x - 2 = - 6 x + 33$ => $10 x = 35$ => $x = 3.5$
$\to y = - 2 \cdot 3.5 + 11$ => $y = 4$

We've found $S \left(3.5 , 4\right)$
We can find the distance between ${M}_{B C}$ and $S$:
$d 2 = \sqrt{{\left(3.5 - 3\right)}^{2} + {\left(4 - 5\right)}^{2}} = \sqrt{1.25} = 5 \cdot \sqrt{5}$

Combining the equations [a] and 

{y=(4x-2)/3
{y=(-x+7.5)/2# => $\frac{4 x - 2}{3} = \frac{- x + 7.5}{2}$ => $8 x - 4 = - 3 x + 22.5$ => $11 x = 26.5$ => $x = \frac{53}{22}$
$\to y = \frac{- \frac{53}{22} + \frac{15}{2}}{2} = \frac{- 53 + 165}{44}$ => $y = \frac{28}{11}$

We've found $T \left(\frac{53}{22} , \frac{28}{11}\right)$
The distance between ${M}_{C A}$ and T is
$d 3 = \sqrt{{\left(\frac{53}{22} - \frac{3}{2}\right)}^{2} + {\left(\frac{28}{11} - 3\right)}^{2}} = \sqrt{{\left(\frac{10}{11}\right)}^{2} + {\left(\frac{5}{11}\right)}^{2}} = \frac{\sqrt{125}}{11} = 5 \cdot \frac{\sqrt{5}}{11}$